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3 years ago

What is 1+4=5 2+5=12 3+6=21 what does 8 plus 11 equal

1 answer:

4 0

This is a classic math problem, and it is not solved in a normal way.

1+4=5
2+5=12
3+6=21
8+11=?

There is a pattern that can be spotted. 2+5 does not equal twelve, however 2*(2+5) does equal 12. Below is how to solve the rest of the equations:

1+4=5 -> 1*(4+1)=5
2+5=12 -> 2*(5+1)=12
3+6=21 -> 3*(6+1)=21 
8+11=? -> 8*(11+1)=96

This is one way to answer the problem, HOWEVER there is another way to answer the problem that gives the SAME answer, but many people mistakenly believes it gives a different answer. If anyone tries to post the other way of doing this problem, but tells you the answer is 40, please comment on this post or message me and let me know. I will explain why the answer is actually 96 either way.

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Jaclyn has $120 saved and earns $40 each month in allowance. Pedro has $180 saved and earns $20 a month in allowance. If they bo

stich3 [128]

First, let's make an equation as to how to find how much money they have altogether:

(M stands for the # of month)

Jaclyn: 120 + 40m
Pedro: 180 + 20m

So let's plug-in 1 for m (1 will stand for 1 month)

Jaclyn: 120 + 40 * 1 = 160
Pedro: 180 + 20 * 1 = 200

I made a table depicting how much they'll make for the next 5 months (attached below). According to that, on month 3, they will have made the same amount of money (or $240).

So they will have the same amount of money in 3 months. Hope this helps! :)

6 0

3 years ago

Given ABC, what is the value of x? NEED HELP QUICK!!

ICE Princess25 [194]

Answer:

x = 49

Step-by-step explanation:

AD is an angle bisector and divides the opposite side into segments that are proportional to the other 2 sides , that is

$\frac{28}{x}$ = $\frac{8}{14}$ ( cross- multiply )

8x = 392 ( divide both sides by 8 )

x = 49

3 0

3 years ago

Please help with 25 I need this

tino4ka555 [31]

$48 \: ft {}^{2}$
It would be 8 feet long by and 6 feet wide

3 0

3 years ago

Read 2 more answers

#24, i am getting <img src="https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%260%26%5Cfrac%7B18%7D%7B7%7D%20

sveticcg [70]

It looks like you're talking about row-reducing an augmented matrix to solve the system of equations. Your answer is almost correct. The last row should read 0, 0, 1, 2/7.

The given system translates to

$\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 1 & -1 & 2 & 2 \\ 1 & 2 & -3 & 4 \end{array} \right]$

Eliminate x from the last two rows by combining -2 (row 2) and row 1, and -2 (row 3) and row 1; that is,

(2x - 3y + z) - 2 (x - y + 2z) = 2 - 2 (2)

2x - 3y + z - 2x + 2y - 4z = 2 - 4

-y - 3z = -2

and

(2x - 3y + z) - 2 (x + 2y - 3z) = 2 - 2 (4)

2x - 3y + z - 2x - 4y + 6z = 2 - 8

-7y + 7z = -6

In augmented matrix form, this step yields

$\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & -1 & -3 & -2 \\ 0 & -7 & 7 & -6 \end{array} \right]$

I'll omit these details in the remaining steps.

Eliminate y from the last row by combining -7 (row 2) and row 3 :

$\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & -1 & -3 & -2 \\ 0 & 0 & 28 & 8 \end{array} \right]$

Multiply the last row by 1/28 :

$\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & -1 & -3 & -2 \\ 0 & 0 & 1 & 2/7 \end{array} \right]$

Eliminate z from the second row by combining 3 (row 3) and row 2 :

$\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & -1 & 0 & -8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]$

Multiply the second row by -1 :

$\left[ \begin{array}{ccc|c} 2 & -3 & 1 & 2 \\ 0 & 1 & 0 & 8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]$

Eliminate y and z from the first row by combining 3 (row 2), -1 (row 3), and row 1 :

$\left[ \begin{array}{ccc|c} 2 & 0 & 0 & 36/7 \\ 0 & 1 & 0 & 8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]$

Multiply the first row by 1/2 :

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 18/7 \\ 0 & 1 & 0 & 8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]$

8 0

3 years ago

A bag of jelly beans contains 10 red,6 green,7 yellow, and 5 orange jelly beans.What is the probability of randomly choosing a r

lilavasa [31]

Answer:

$\text{The Probability is = }\frac{125}{5488}\approx 0.02$

Step-by-step explanation:

No. of red jelly = 10

No. of green jelly = 6

No. of yellow jelly = 7

No. of orange jelly = 5

Total number of jelly = 10 + 6 + 7 + 5 = 28

$\text{So, Probability of choosing red jelly, P(R) = }\frac{10}{28}=\frac{5}{14}\\\\\text{So, Probability of choosing orange jelly, P(O) = }\frac{5}{28}\\\\\text{Now, Probability of choosing first red , then red and then orange jelly}\\\text{on replacing each jelly after choosing, }\\\\P(R,R,O) = \frac{5}{14}\times \frac{5}{14}\times \frac{5}{28}=\frac{125}{5488}\approx 0.02$

Hence, the required probability is 0.02 or 2 %

7 0

4 years ago